I spent a long time typing up an involved reply. A few minutes before finishing it off, my computer reset itself. So I'm not going to attempt a response again until I get home after work, which may be very late in the day or even tomorrow (or later). For now:

There is no unique combination that deductively follows. All could be knaves, for example. There are other combinations that work, too.

Remember knaves alternately tell the truth and lie.

After their first response they're deterministic.

Hint: use Notepad and save often, then copy into your reply.

The Eleven Gemstones

(This is a very easy problem, IMO, but it's fun, at least I think so)

The Mine Overlord passes away, leaving his 11 precious gemstones that he has mined over the years to the Chief-Miner, the Sub-Chief-Miner and the Minor-Miner-Squadron-Leader.

He declares that the Chief-Miner, being the highest ranked, shall receive half of the gemstones. The Sub-Chief-Miner, the next in rank, shall receive a third of the gemstones, and the Minor-Miner-Squadron-Leader shall receive a twelfth. Anything left over, if there is anything, goes to charity.

How can they divide the eleven precious gemstones into halves, thirds and twelfths??? Frustrated, they call the famous Jewel Mathematician, who comes with her priceless gemstone brooch. "Easy," she says, adding her own gemstone to the pile. "There are twelve. Chief-Miner, you take 6, which is half. Sub-Chief-Miner, you take 4, which is a third. Minor-Miner-Squadron-Leader, you take 1, which is a twelfth. That's the 11 gemstones, and I'll take my own gemstone brooch back."

She then departs, leaving them confused, yet divided up nicely, with her still being able to keep her gemstone brooch.

***

What is the hole here?

1/2 + 1/3 + 1/12 = 11/12.

Adding a stone to get 12 permits the 11 to be divided in the right proportion and leaves the 12th stone undistributed.

This also works with 17 using 1/2, 1/3 and 1/9 [17/18]

and with 19 using 1/2, 1/4 and 1/5 [19/20].

In the weird land of KKL, Knights always tell the truth, Knaves alternately tell the truth and lie and Liars always lie.

You have been sent on assignment to KKL to learn the ways of the natives and explore ways of identifying their type.

A guide leads you to a dark alley and introduces you to Al, Bob and Chuck, who agree to answer exactly three questions.

It's too dark to see their faces, you don't know which type of citizen each is, you can't tell which of them is speaking, and they reply in random order. Although they answer anonymously, you must determine which type of citizen they are.

You decide to start with the direct approach and then play it by ear:

[1] I'd like to know what type of citizens I'm talking with. Please tell me.

Sure thing. Myself, I'd be misleading you if I didn't tell you I'm a knight.

Hate to admit it, but I'm the unreliable one, being as how I'm a knave.

Don't listen to anything I tell you mate, I'm a bloody liar.

[2] OK great, thanks for that. But I wonder whether there really is one of each type.

Maybe there's more than one of some type. Can you comment specifically on that?

Pretty shrewed, aren't you? Well, you're right - it turns out we're all the same type.

Forget him. We told you already: there's exactly one of each type here.

Aww, they're both messing with you: actually, there's exactly two of one type here.

[3] OK well that's pretty confusing. For a final question, it might help just to talk about Al.

Would you tell me what type of citizen Al is? I think that will tell me all I need to know.

Why all the attention for Al? He's just a knave.

No way! Al is a liar!

Well I'll tell you this much, whatever Al is, Bert Bob is something else.

OK guys, thanks a lot.

I think now I know what type Chuck is.

And let's see ... what type Al is ... and also Bob.

Are you messing with us, or do you really know?

Yeah, I assumed pouring 1/2 the mix not 1/2 liter.

I thought of seeing what happened if you do that.

But if you don't pour 1/2 liter each time, the jars start containing unequal amounts of liquid after a set of pours.

I didn't think there'd be a chance of equal distribution of liquids in that case.

users can edit their posts only within 10 minutes after they post ... so that we can keep track of changes (and nobody is hiding anything)

That's usually handled automatically by the software, which places an edit history

[last edited on ___ ; ___ edits total]

at the end of the post if another post is made before the edit.

The editor can then append a note saying "Edited to clarify the pouring rules," or some such.

I suppose there is some merit to locking after 10 minutes,

but it does have the drawback of keeping things from being fixed

or, say, withdrawing a spoiler, etc. without forum owner's intervention.

[Dividing the gems]

Use abbreviations to simplify.

H = Haxite; A = Aquadium.

A = 3H.

They mined 14A + 26H = 14[3H] + 26H = 68H.

68 is not a nice number.

So add theChief-Miner's Haxite necklace = 69H.

Each prince should get 23H.

23H = 4A + 11H = 5A + 8H.

Prince 1 gets 4A + 11H

Prince 2 gets 5A + 8H

Prince 3 gets 5A + 8H.

Chief Miner gets shafted, but lives to mine another day.

You have three 2-liter jars, call them 1, 2 & 3. Each jar has 1 liter of a different liquid in it - so:

Jar 1 - has liquid A

Jar 2 - has liquid B

Jar 3 - has liquid C.

Your task, pour 1/2 liter from Jar 1 into Jar 2. Now pour 1/2 liter of the mix from Jar 2 to Jar 3.

Now pour 1/2 of the mix from Jar 3 back into Jar 1.

Assume when you pour, you pour exactly 1/2 liter and the liquid mixes perfectly each time.

Question - how often must you repeat the round of pourings (1 to 2, 2 to 3, 3 back to 1) until each jar holds EXACTLY 1/3 of each of the original liquids?

Clarify: you pour 1/2 liter from Jar3 to Jar1, not 1/2 of Jar3's mix, which would be 3/4 of a liter.

After the 48th pour Jar 1 has exactly 1/3 liter of liquid A

After the 49th pour Jar 2 has exactly 1/3 liter of liquid B

After the 50th pour Jar 3 has exactly 1/3 liter of liquid C

Where "exactly" means within an error bound of about 10**-16, the precision of my processor.

If after some pour the exact conditions obtain, I'd need a different method to find it.

By the way, are we actually counting molecules here? B)

If you throw a dice once the cance of getting six is 1/6

If you throw a dice twice the cance of getting six is 2/6

If you throw a dice three times the cance of getting six is 3/6

If you throw a dice four times the cance of getting six is 4/6

If you throw a dice five times the cance of getting six is 5/6

Does this mean that if you throw a dice 6 times you are garantied 6?

Of course it doesn't but how come theareticly the chance is 100% when practicly it isn't.

If you throw two dice there are 36 outcomes.

Of those, only eleven outcomes show a six.

Namely, [16] [26] [36] [46] [56] [66] [65] [64] [63] [62] [61]

Your chances of getting a 6 with two dice are therefore 11/36; not 12/36.

That's the brute force way to analyze, and it will always work for you.

[1] find out how many equally likely ways [N] things could turn out.

[2] count the outcomes [F] that are favorable.

[3] your chances are F/N.

There's an easier way to do this.

Each die has a 5/6 chance of NOT being a six.

Throwing two dice gives you a [5/6][5/6] = 25/36 chance of neither of them being a 6.

The chance that one or both are a 6 is therefore 1-25/36 = 11/36.

What if you throw six dice?

Your chances of getting at least one six are about 2/3.

1-[5/6][5/6][5/6][5/6][5/6][5/6] = 0.6651020233

[3 Doors and 2 Goats]

This is a classic puzzle, most recognizably called the Monty Hall problem.

It's been posted elsewhere in this discussion board.

Your chosen door has 1/3 chance of getting you a car.

The exposed goat door has a 0/3 chance of getting you a car.

The third door thus has a 2/3 chance of getting you a car.

You should swap.

[3 Princess Puzzle]

I don't think 3 or 4 work.

3) Okay, so if I ask Honest (A) if B is more likely to lie than her, she says yes.

But, she says yes whether B is Mischievous or Liar, so I may end up with Mischievous,

never mind the fact that I don't know that A is honest.

The case is the same with Liar.

She will say yes to either one so there is no way to distinguish between them.

So basically, if you ask the question of Liar or Honest, you have a 50% chance.

You're right - [3] is a broken version of [2].

I had something else in mind. Good catch..

4) Okay, so I ask A if B is Mischievous. She says yes.

I don't know whether A is Honest or Liar (or Misch but that point is moot since you're a winner either way if you ask her), so do I pick B or C?

So, again, my future lies on chance. Not cool.

Actually, [4] works. Remember, I said ...

In each case, if the answer is Yes choose C; if No, choose B.

PT and PL will both reliably identify PM.

If A is PT, she tells the truth about telling the truth, and if A ls PL, she lies about lying.

Way to go Ploper!

------+----+----+----+----+
| MB | PL | LF | TG |
------+----+----+----+----+
Al | | X | | X |
Bill | X | | | |
Chuck | | | X | X |
------+----+----+----+----+
right | 2 | 2 | 2 | 1 |
------+----+----+----+----+

[Hats on a death row!! One of my favorites puzzles!]

First guy is a coin toss - let's wish him good luck.

His job is to establish the parity of black hats visible to him.

He says "Black" if he sees an odd number of black hats; "Red" otherwise.

By paying attention to what has been said, each prisoner will know his hat's color.

Example:

Second to speak hears "Black" and sees an even number of black hats.

He knows his hat is black [odd changed to even - must be his is black] and says "black".

Third guy has heard "black" and "black" and sees an even number of black hats.

He knows his hat is red [even stayed even - his hat can't be black] and says "red".

And so on, to the front of the line.

General algorithm:

The first time you hear "black", say to yourself "odd".

Each time your hear "black" after that, change the parity: "even", "odd", ... etc.

When it's your turn, if the black hats you see match the running parity, you're Red; Black otherwise.

Call out your color.</div></div></div>

[Hats on a death row!! One of my favorites puzzles!]

I hope its not something stupid like "Oh to make it a fair challenge the evil king makes everyone's hat color the same", I highly doubt it.

It's not. The strategy works for any distribution of colors.

Is my answer on the right track at least?

Yes. The survival of the first prisoner to speak cannot be guaranteed.

The others, if they follow the strategy and listen to all of the previous speakers, will know their hat color.

Roolstar gave a good clue:

The most reliable and correct method should devide the big universe of possibilities of the 20 prisoners into 2 possibilities

The first speaker tells which of these possibilities is true. What are the possibilities?

[3 Princess Puzzle]

I believe there are at least three questions that work.

Here are four questions that work.

Select the least desirable Princess and call her A.

[1] is B older than C?

[2] does B lie less frequently than C?

[3] is B more likely to lie than you?

[4] if I were to ask, would you say that B is Princess Mischievous?

In each case, if the answer is Yes choose C; if No, choose B.

Princess Truth and Princess Lie will answer Yes if Princess C is the other of these two, and No if Princess B is.

If A is Princess Mischievous, a random B or C choice gets you a winner either way.

[Hats on a death row!! One of my favorites puzzles!]

1

[3 Princess Puzzle]

The clunker in this riddle is Princess Mischievous.

You might, unknowingly, ask your question of her, and her answer is useless!

Devise a strategy whereby after asking one of the Princesses a question

you will pick one of the other two as a wife. That way, if you happen

to ask Princess Mischievous the question, her answer won't matter.

Either of the others is an acceptable choice!

All that's needed is a question which, if answered Yes [or No] by either

Princess Truth or Princess Lie, will identify Princess Mischievous.

A Yes/No question can guide a binary choice.

To optimize the strategy:

Since you will not choose her, ask the question of the ugliest of the three.

[3 Princess Puzzle]

There are 3 unknowns, and a Yes/No question discriminates between 2 and only 2 possibilities.

You don't need a complete solution.

Your choice either is or is not the mischievous one - two cases.

I believe there are at least three questions that work.

[Favorite Riddle of All Time]

[tongue placed firmly in cheek]

Well it would have to be one from the Alex and Morty's series; modesty prevents me from choosing one.

[tongue removed from cheek]

[Death Probability! Really Hard One!]

[1] Alfred will survive - 1/3 chance; or [2] Bob will survive - 2/3 chance.

I don't see why that's [a] incorrect, or incomplete.

[easy puzzles..]

ESCAPE of MARTHA

If I were to ask you which door leads to freedom, which door would you point to? Believe the answer.

THE WATER PROBLEM --

Pour the first glass into the pail. Freeze it. Pour the second glass into the pail

LOCKED with LOCKERS

1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400 441 484 529 576 625 676 729 784 841 900 961.

The others are open

[Matchsticks]

I am very confused???? the solution to what?

The first arrangement can be read: the square root of 1 equals 11.

The second arrangement [inside the spoiler] can be read: 11 divided by 1 equals 11

The puzzle asks you to move two matchsticks to make a valid equation.

OK?

[Death Probability! Really Hard One!]

Possibility One: Alfred and the first of the other two get executed.

Possibility Two: Alfred and the second of the other two get executed.

Possibility Three: Both of the other two get executed.

As we see, Alfred has 2 chances in 3 of being executed.

Now, Alfred asks the guard to name which of the other two is being executed (or to pick one of the other two, if both are to be executed).

But Alfred already knows that at least one of them is to be executed!

Finding out the name of the victim doesn't give him any relevant new information. [* - see below]

To put it another way: Alfred doesn't know whether Charlie is "the first" or "the second" of the "other two" in the possibilities listed above.

The exact same three possibilities still exist, even after Alfred gets a name out of the guard.

Agree.

Alfred knows - and we know - either

[1] Charlie is "the first" - in which case Possibility Two has been eliminated - i.e. determined to be false. or

[2] Charlie is "the second" - in which case Possibility One has been eliminated - i.e. determined to be false..

Can you clarify why you say a possibility has not been eliminated?

"The exact same three possibilities still exist ... "

An option that has a zero fraction of the total probability [i.e. is false] is usually thought of as having been eliminated.

The remaining possibilities - the options with non-zero probabilities - are:

[1] Charlie and Alfred die - probability is 2/3.

[2] Charlie and Bob die - probability is 1/3.

If you want to hang onto your third possibility, then:

[3] Albert and Bob die - probability is 0/3.

[* - you can argue it's not relevant to Alfred, but it's certainly relevant to Bob.]

Since Bob could have been named, but was not, Bob's chances of survival just doubled.

Alfred's chances remained the same.

Proof:

Alfred's chances [of survival] stayed at 1/3.

Charlie's chances went to zero.

Bob's doubled, to 2/3.

Since one and only one survives, these chances must sum to unity.

This is why in Monty Hall you make the swap.

[Need name of this puzzle]

Crack Infosys,Wipro,Satyam,TCS Exams by solving Puzzles,You can solve puzzles by preparing Shakuntala Devi Puzzle solving ebook.You can download Shakuntala Devi Puzzle Solving Ebook from

[url deleted by admin]

Put things like that here:

http://brainden.com/forum/viewforum.php?f=9

[Alex's gigantic mug puzzle]

Edited out a rant on some rogue user for spamming.

Where did I put that eraser?

[IMPOSSIBLE PUZZLE]

I know a puzzle which NO one has been able to figure out so if you think you can then please do!! But I don't know how to put it on here cause it needs to be drawn could someone please tell me? It looks sorta like this-

l---*--- l---*---- l   Ok thats very messy but you have to go through every line only once and only with one line cause you 

*		  *			*	can't take you pen of the page you have to get one line through each of those lines once. 

l		   l			 l   Think you can  do it? It's harder then you think. I have marked each line you have to go through with a 

l-*-l-*--l--*--l-*-l   star*!!

*	*			*   l

l	 l			  l   *

l-*-l----*---- l-*-l[/code]
    [b]xOzoeOx[/b], just use "Code" around that part of the text to get a monospaced font. 

Then you can line things up ... ok?